Mathematics

Ryan Unger

4:00 PM

that's what you actually use most of the time

Balarka Sen

inequalities? what are they?

i only do topos theory thank you

Ryan Unger

something like that

Semiclassical

There's a great pair of quotes I saw juxtaposed once and haven't been able to find them since

1) "Every nontrivial equality is the result of two inequalities."
2) "To deduce an equality from two inequalities is to learn nothing."
(paraphrased)

I'm not doing them justice tho

I want to say one of them was by Noether?

MatheinBoulomenos

@user170039 if $J(R(G))=0$, then your argument might actually work, I think, since then the map $R(G) \to \prod_{\text{$S$ simple $R(G)$-module}}S$ is injective. simple $R(G)$-modules are of the form $S=R(G)/M$ for $M$ a maximal left ideal. As $M$ is a maximal subgroup by your subgroup/ideal correspondence, $R(G)/M$ is a simple abelian group

Alessandro Codenotti

@RyanUnger Exercise 3.32 in Brezis

MatheinBoulomenos

4:04 PM

@RyanUnger my eyes

Alessandro Codenotti

It says that in a uniformly convex Banach space projections onto a closed convex set exist and are unique

bolbteppa

If you stand on some train tracks and stare head on at them, then step left and keep staring at the train tracks, is the image you see then the action of a transformation in the isometry group of projective space $\mathbb{R} \mathbb{P}^3$

Semiclassical

(finding the source of those quotes bugs me every time I remember them, and I never manage it)

Alessandro Codenotti

I'm not sure but I think reflexive Banach space might be enough, is the first proof of 5.2 (also in Brezis) using more properties of Hilbert spaces or just reflexivity?

Semiclassical

ahah! books.google.com/…

In algebra, when one says a = b, it is a tautology and so uninteresting; while in analysis, when one says a = b, it is two deep inequalities. —attributed to S. Bochner

If one only proves a = b by showing a ≤ b and b ≤ a, one has not understood the true reason why a=b.
—attributed to E. Noether

MatheinBoulomenos

4:08 PM

@user170039 nevermind, $R(G)$ is always a radical ring

Balarka Sen

@bolbteppa Uh. I mean, I guess you can say that.

Semiclassical

I always liked the contrast between those two views

bolbteppa

Cool

Balarka Sen

Isometries of $\Bbb{RP}^n$ aren't complicated, they all come from isometries of the universal cover $S^n$. Indeed, it's $SO(n+1)/\{\pm I\}$

bolbteppa

Right but relating that to DaVinci paintings is the hard part ;)

Balarka Sen

4:10 PM

So if you have a pair of lines in $\Bbb{RP}^n$, under some isometry they get mapped into some other pair of lines. This is the same as the situation with two great circles in $S^n$ (indeed, there's a "commutative diagram" of these four configurations)

MatheinBoulomenos

actually there's an algebraic way to prove inequalities for integers: prove divisibility relations, there are some deep and difficult proofs in NT (namely CFT) that prove a=b by proving that a divides b and b divides a

if you use the analytic proofs for the first and second inequalities in CFT you get actual inequalities and if you use the algebraic proofs you get divisibility relations

bolbteppa

The isometries of Euclidean space are translations, rotations, reflections and combinations of them - can you think of the isometries of $\mathbb{R} \mathbb{P}^n$ somewhat similarly?

Balarka Sen

Isometries of $S^n$ are generated by rotations about codimension $2$ planes and reflections. These project to isometries of $\Bbb{RP}^n$.

Semiclassical

@MatheinBoulomenos yeah, and I'd count that as still being in the spirit of those quotes. just a different partial ordering than usual

Balarka Sen

You shouldn't think of $\Bbb{RP}^n$ as the stratified $\Bbb R^n \cup \Bbb{RP}^{n-1}$ when talking about isometries; that's not the right picture. The top stratum $\Bbb R^n$ there is not Euclidean.

MatheinBoulomenos

4:13 PM

@Semiclassical I was referring to the "in algebra" and "in analysis" part of the first quote

Semiclassical

ahhh

@RyanUnger reason I was wondering was because of an old statistics paper (1937) where they set up the notion of a vector space with inner product (not quite the L^2 case I gave, but closely related) but don't talk about finite second moment. But it's a rather informal paper, so I suspect that omission isn't to be taken seriously

(or "it's obvious that you exclude random variables with infinite variance, since this ensures that the inner product is well-defined in the first place")

bolbteppa

[hmm](https://en.wikipedia.org/wiki/Projective_orthogonal_group#Odd_and_even_dimensions)
"In odd dimension, $SO ( 2 k + 1 ) ≅ P S O ( 2 k + 1 ) = P O ( 2 k + 1 ) , {\displaystyle SO(2k+1)\cong PSO(2k+1)=PO(2k+1),} SO(2k+1)\cong PSO(2k+1)=PO(2k+1)$, so the group of projective isometries can be identified with the group of rotational isometries.

In even dimension, $SO(2k) → PSO(2k)$ and $O(2k) → PO(2k)$ are both 2-to-1 covers, and $PSO(2k) < PO(2k)$ is an index 2 subgroup. " seems you need to be more careful interpreting the isometries in projective space

MatheinBoulomenos

@BalarkaSen @bolbteppa me being an algebraist, I'd want to emphasize that the reason every isometry of $S^n$ projects nicely onto $\Bbb R \Bbb P^n$ is that $\Bbb{RP}^n$ is the quotient of a central element in the isometry group of $S^n$, namely $-I$. If you quotient by a non-central subgroup of the isometry group (e.g. in a lens space), then only elements in the centralizer of that subgroup project to isometries on the quotient

Semiclassical

(The slightly annoying thing for purposes of describing the paper is that it's not quite "L^2 w/r/t probability measure" in the way I just said. Instead they take the equivalence relation to be "equality a.e. up to a real constant" and the inner product to be E[XY]-E[X]E[Y]. Still a Hilbert space but a quotient of L^2.)

Balarka Sen

@MatheinBoulomenos $\Bbb Z_2$-equivariant isometries of $S^n$ as the same as the isometries of $S^n$, sure.

MatheinBoulomenos

4:23 PM

yeah, that was my point, I know it's trivial, but only a posteriori if you know that isometries are linear

Semiclassical

Not sure how to notate that, come to think of it. $L^2(\text{probability space})/\mathbb{R}$?

MatheinBoulomenos

@bolbteppa projection gives a surjective group homomorphism $SO(n) \to PSO(n)$, so you have to compute the kernel. If $A$ is an isometry of $S^n$ that acts trivially on $\Bbb R \Bbb P^n$, this means that $Av=\pm v$ for all $v \in S^n$, by continuity, the sign is constant, so $A=\pm I$, but if $n$ is odd, then $-I \notin SO(n)$

bolbteppa

Hmm, right

MatheinBoulomenos

of course, you have to specify if you care for orientation or not

Semiclassical

In another matter: I've heard the name Kullback before, namely with reference to the Kullback-Liebler divergence. (not that I know what that is, but I've heard that phrase before)

what I hadn't connected until recently was that he was also an important WW2 cryptanalyst

I knew Turing was (ofc) but there's other important names in math which passed through there

famesyasd

4:34 PM

Can anyone tell me why is it obvious that the result of integration does not depend on a cell we are choosing? I think I can rpove this but it would require several lemma's and this is far from being obvious.

Semiclassical

test: $\supp{f}$

nuts

note that $\operatorname{supp}f\subset I^k$. Then $\int_{I^k} f =\int_{\operatorname{supp}f} f+\int_{I^k\setminus \operatorname{supp}f}=\int_{\operatorname{supp}f}$

is there anything wrong with that? feels like I may be making things too easy

famesyasd

yeah I don't have that property

about additivty

Semiclassical

Hmm

famesyasd

the only thing I have is a definition of an integral on a cell and that it does not depend on the order of integration. That's it and the page that I've linked

I thought of taking two cells, and making out of them a third one like this:

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$Mathematics$ Mathematics